# CBSE Class 12th Math 13 – Probability MCQs

#### The probability that a shooter can shoot the target is 0.9. The probability that out of 10 shots, he can shoot at least one is ____.

Let X denote the number correct shots in 10 shots. The probability to shoot correctly remains the same for every shot. Clearly, X has the binomial distribution with n = 10 and probability for success p = 0.9 and probability of failure q = 0.1. From Bernoulli distribution, P(X = x) = C(n, x) × p^{x }× q^{(n - x)}, where x = 0, 1, 2,..., n. Here n = 10. P(X >= 1) = 1 - P(X = 0) = 1 - [C (10, 0) × (0. 9)^{0} × (0.1)^{(10-0)}] =1- (1/10)^{10} = 1 - (10^{-10}) Hence, the probability that out of 10 shots, he can shoot at least one is 1 - (10^{-10}).

#### In a Bernoulli trial, the probability of success is m and probability of failure is n. Then, the arithmetic mean of the probability of success and failure is ________.

In Bernoulli trials, the probability of success + probability of failure = 1 ⇒ m + n = 1. Now, Arithmetic mean = (m+n)/2 = 1/2 (Since from equation, m + n = 1) Hence, the arithmetic mean of probability of success and failure is 1/2.

#### In Bernoulli trials, the number of outcomes is ____.

In Bernoulli trials, each trial has exactly two outcomes: success or failure. Hence, in Bernoulli trials, the number of outcomes is 2.

#### Out of 10 boys of class XII of a school, 6 students were in 1^{st} Division, 3 students were in 2^{nd}Division and one student failed in the last year's exam. If three students are selected with replacement, the probability that at most two of them are in the 2^{nd }division is ____.

Let X denotes the number of 2^{nd } division students selected out of 10 students. Clearly, X has the binomial distribution with n = 10 and probability for success p = 0.3 and probability of failure q = 0.7. From Bernoulli distribution, P(X = x) = C(n, x) × p^{x} × q^{(n-x)}, where x = 0, 1, 2,..., n. Here, n = 3. Probability that at most two of them are in 2^{nd} division is P(X ⇐ 2) = P(X = 0) + P(X = 1) + P(X = 2) P(X ⇐ 2) = [C(3, 0) × (0.3)0 × (0.7)(3-0)] + [[C(3, 1) × (0.3)1 × (0.7)(3-1)] + [C(3, 2) × (0.3)2 × (0.7)(3-2)] = 0.973 Hence, the probability that at most, two of them are in 2^{nd } division is 0.973.

#### In Bernoulli trials, if the probability of success is p in the first trial, then the probability of success in the 5^{th} trial for the same experiment is ____.

By the definition of Bernoulli trials, the probability of success remains the same in each trial. Hence, the probability of success in the 5^{th} trial for the same experiment is p.

#### A bag contains 2 white and 1 red balls. One ball is drawn at random and then put back in the box after noting its colour. The process is repeated again. If X is a random variable that denotes the number of red balls recorded in the two draws, the average of the possible different values of X is ____.

Let the balls in the bag be denoted by w_{1}, w_{2}, r, where w stands for white ball and r for red ball. Then, the sample space is S = { w_{1} w_{1}, w_{1} w_{2}, w_{2} w_{2}, w_{2} w_{1}, w_{1} r, w_{2 }r, r w_{1}, r w_{2}, r r} Now, for K∈S, X(K) = number of red balls Therefore, X({w_{1} w_{1}}) = X({w_{1} w_{2}}) = X ({w_{2} w_{2}}) = X({w_{2} w_{1}}) = 0 X({w_{1} r}) = X({w_{2} r}) = X ({r w_{1}}) = X ({r w_{2}}) = 1 and X ({r r}) = 2 Thus, X is a random variable which can take values 0, 1 or 2. And average of 0, 1 and 2 is (0 + 1 + 2)/3 = 1 The average of the possible different values of X is 1.

#### If a random variable X represents the number of heads obtained when a coin is tossed 2 times, the maximum value of X is ____.

Let us consider the experiment of tossing a coin two times in succession. The sample space of the experiment is S = {HH, HT, TH, TT}. If X denotes the number of heads obtained, then X is a random variable and for each outcome, its value is as given below: X(HH) = 2 X(HT) = 1 X(TH) = 1 X(TT) = 0 Thus, X is a random variables and which can take values 0,1 or 2 And the maximum value of X is 2 Hence, the maximum value of X is 2.

#### Hence, the marks obtained by the student for this question is 3. (i) Two balls are drawn successively from an urn containing 2 black and 3 white balls, with replacement. (ii) Two balls are drawn successively from an urn containing 2 black and 3 white balls, without replacement. (iii) A fair coin is tossed 5 times, where the outcome as head is a success. If the interviewer gives 3 point for each correct example and deducts 1 point for the wrong example, the points obtained by the student for this answer is ____.

As per the definition of Bernoulli trials, example number (ii) two balls are drawn successively from an urn containing 2 black and 3 white balls without replacement – is not a Bernoulli trials. This is because, in the first trial, the probability of success (say black) is 2/5, while in the 2^{nd} trial, the probability of success (say black) is 1/4 if black is drawn in the first trial, or 2/4 if white is drawn in the first trial. So, the probability of success is not the same in each trial. Hence, it is not an example of Bernoulli trials. The remaining two examples are correct examples of Bernoulli trials. The points obtained by the candidate for 2 correct & 1 wrong example is, (2 × 3) – (1 × 1) = 5. Hence, the student will get 5 points for this answer.

#### Sachin can hit 2 sixes in 10 balls. The probability that Sachin can hit 2 sixes in an over is ____.

Let X denotes the number of sixes hit by Sachin. For each ball, the probability to hit six is same. Clearly, X has the binomial distribution with n = 6 and probability for success p = 0.2 and probability of failure q = 0.8. From Bernoulli distribution, P(X = x) = C(n, x) × p^{x }× q^{(n - x)}, where x = 0, 1, 2,..., n. Here, n = 6. P(X = 2) = [C(6, 2) × (0.2)^{2} × (0.8)^{(6-2)}] = 0.25 (approx.) = 0.24576 Hence, the probability that Sachin can hit 2 sixes in an over is 0.25.

#### The probability that a building built by ABC company will collapse after 300 years is 0.4. The probability that out of 10 buildings built by the same company, all buildings will be safe after 300 years is ____.

Let X denote the number of buildings that can collapse out of 10 buildings, after 300 years. From Bernoulli distribution, P(X = x) = C(n, x) × p^{x} × q^{(n-x}), where x = 0, 1, 2,..., n. Here, n = 10, P(X = 0) = C(10, 0) × (0.4)^{0 }× (0.6)^{(10-0)} = (0. 6)^{10} The probability that out of 10 buildings built by the same company, all buildings will be safe after 300 years is (0. 6)^{10}.

#### In class XII examination, a student has written the following four conditions in favour of Bernoulli trials: (i) There should be a finite number of trials. (ii) The trials should be dependent. (iii) Each trial has exactly two outcomes: success or failure. (iv) The probability of success remains the same in each trial. If the question was of 4-marks and teacher gave one mark for each correct condition, the marks obtained by the student for this question is ____.

By definition, trials of a random experiment are called Bernoulli trials, if they satisfy the following conditions: (i) There should be a finite number of trials. (ii) The trials should be independent. (iii) Each trial has exactly two outcomes: success or failure. (iv) The probability of success remains the same in each trial. The condition (ii) is “The trials should be independent”, however the student have written “dependent”, which is wrong. The other three conditions are correct. Hence, the marks obtained by the student for this question is 3.

#### If a random variable X represents the number of heads obtained when a coin is tossed 2 times, the prime number value of X is ____.

Let us consider the experiment of tossing a coin two times in succession. The sample space of the experiment is S = {HH, HT, TH, TT}. If X denotes the number of heads obtained, then X is a random variable and for each outcome, its value is as given below: X(HH) = 2 X(HT) = 1 X(TH) = 1 X(TT) = 0. Thus, X is random variables and can take values 0, 1 or 2. And, the Prime number value of X is 2.

#### In a village, the probability that a person is not an engineer is 0.8. If two people are selected with replacement, then the probability that out of 3 people, two are engineers is ____.

Let X denote the number of engineers in an experiment of 3 trials. Clearly, X has the binomial distribution with n = 3 and probability for success p = 0.2 and probability of failure is q = 0.8 Now, from Bernoulli distribution, P(X = x) = C(n, x) × p^{x} × q^{(n-x)}, where x = 0, 1, 2, ..., n Here n = 3, P(X = 2) = C(3, 2) × (0.2)^{2} × (0.8)^{(3-2)} = 3!/(2!1!) × (0. 04) × (0. 8) = 0.096 Hence, the probability that out of 3 people, two are engineers is 0.096.

#### A person plays one lottery every day for a week and the chance of winning is 0.1 in a day. The probability that he can win exactly six days in the week is _____ ÷ 1000000.

Let X denotes the number of winning days, in an experiment of 7 trials. Clearly, X has the binomial distribution with n = 7 and probability for success p = 0.1 and probability of failure is q = 0.9 From Bernoulli distribution, P(X = x) = C(n, x) × p^{x} × q^{(n-x)} , where x = 0, 1, 2,..., n. Here n = 7, P(X = 6) = C(7, 6) × (0.1) ^{6} × (0.9)^{(7-6}) = {7!/(6!1!)} × (0. 1) × (0. 9) = 0.0000063 = 6.3 ÷ 1000000 Hence, the probability that he can win exactly six days in the week is 6.3 ÷ 1000000.

#### Let Y represent a random variable which is the difference between the number of heads and the number of tails obtained when a coin is tossed 2 times. The number of different values of Y is ____.

Let us consider the experiment of tossing a coin two times in succession. The sample space of the experiment is S = {HH, HT, TH, TT}. Let Y denote the number of heads minus the number of tails for each outcome of the above sample space S. Then, Y(HH) = 2, Y (HT) = 0, Y (TH) = 0, Y (TT) = – 2. The number of different values of Y is 3 i.e. 0, 2 or -2. Hence, the number of different values of Y is 3.